- #1
EinsteinCross
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Let ## \mathbf{x''} = A\mathbf{x} ## be a homogenous second order system of linear differential equations where
##
A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
## and ##
\mathbf{x} = \begin{bmatrix}
x(t)\\ y(t))
\end{bmatrix}
##
Now to solve this equation we transform it into a 4x4 first order system ##\mathbf{X'} = M\mathbf{X} ## where
##
M = \begin{bmatrix}
0 & 1 & 0 & 0 \\
a & 0 & b & 0\\
0 & 0 & 0 & 1 \\
c & 0 & d & 0
\end{bmatrix}## and ##
\mathbf{X} = \begin{bmatrix}
x(t)\\ x'(t)
\\ y(t)
\\ y'(t)
\end{bmatrix}
##
Now solving the first order system by calculating the eigenvalues and corresponding eigenvectors gives us the fundamental matrix for the 4x4 system
##
\Phi (t) = [[\xi_{1}],[\xi_{2}],[\xi_{3}],[\xi_{4}]]
\begin{bmatrix}
e^{\lambda_{1}t }
\\ e^{\lambda_{2}t }
\\ e^{\lambda_{3}t }
\\ e^{\lambda_{4}t }
\end{bmatrix}^{T} ## where the xi brackets are the eigenvectors in column form.
But my question is this: What I am looking to do is to find the fundamental matrix of the original second order 2x2 system which is represented as ##
\Psi (t) = \begin{bmatrix}
\psi_{1}(t) & \psi_{2}(t)
\\ \psi_{3}(t)
& \psi_{4}(t)
\end{bmatrix} ## such that ##\Psi ''(t) = A\Psi (t)##. So how does one extract the solution(s) to the original 2x2 system from the fundamental matrix of the first order 4x4 system?
##
A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
## and ##
\mathbf{x} = \begin{bmatrix}
x(t)\\ y(t))
\end{bmatrix}
##
Now to solve this equation we transform it into a 4x4 first order system ##\mathbf{X'} = M\mathbf{X} ## where
##
M = \begin{bmatrix}
0 & 1 & 0 & 0 \\
a & 0 & b & 0\\
0 & 0 & 0 & 1 \\
c & 0 & d & 0
\end{bmatrix}## and ##
\mathbf{X} = \begin{bmatrix}
x(t)\\ x'(t)
\\ y(t)
\\ y'(t)
\end{bmatrix}
##
Now solving the first order system by calculating the eigenvalues and corresponding eigenvectors gives us the fundamental matrix for the 4x4 system
##
\Phi (t) = [[\xi_{1}],[\xi_{2}],[\xi_{3}],[\xi_{4}]]
\begin{bmatrix}
e^{\lambda_{1}t }
\\ e^{\lambda_{2}t }
\\ e^{\lambda_{3}t }
\\ e^{\lambda_{4}t }
\end{bmatrix}^{T} ## where the xi brackets are the eigenvectors in column form.
But my question is this: What I am looking to do is to find the fundamental matrix of the original second order 2x2 system which is represented as ##
\Psi (t) = \begin{bmatrix}
\psi_{1}(t) & \psi_{2}(t)
\\ \psi_{3}(t)
& \psi_{4}(t)
\end{bmatrix} ## such that ##\Psi ''(t) = A\Psi (t)##. So how does one extract the solution(s) to the original 2x2 system from the fundamental matrix of the first order 4x4 system?